March 17th, 2018, 05:00 AM  #1 
Newbie Joined: Feb 2018 From: Iran Posts: 16 Thanks: 3  Projectile motion A man is standing on a railway car which moves at a constant velocity of 5m/s He releases an object in his hand from a height of 1.8m as shown in the figure; at the same instant the man releases the object, the railway car start slowing down at a deceleration of 1m/s^2. At what horizontal distance from point L on the car does the object strike the car? Please explain step by step. Last edited by skipjack; March 17th, 2018 at 11:52 PM. 
March 17th, 2018, 06:29 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,738 Thanks: 1388 
time it takes for the object to hit the surface of the cart ... $1.8 = 4.9t^2 \implies t = \dfrac{3\sqrt{2}}{7}$ horizontal displacement of the object ... $x_1 = 5t$ horizontal displacement of point L ... $x_2 = 5t  \dfrac{t^2}{2}$ can you finish? 
March 17th, 2018, 09:45 PM  #3 
Newbie Joined: Feb 2018 From: Iran Posts: 16 Thanks: 3  The object is freely falling where did the horizontal displacement of object come from?

March 18th, 2018, 03:45 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,738 Thanks: 1388 
Free fall does not mean an object moves strictly in a vertical direction. It will do so only if it has a zero velocity component in the horizontal direction. $x = x_0 + v_{0x} t  \dfrac{1}{2}a_x t^2$ Both object and position L on the cart’s surface start at $x_0=0$ and both move with an initial horizontal velocity $v_{0x}= 5 \, m/s$ The object’s horizontal velocity remains a constant 5 m/s. It undergoes no acceleration in the horizontal direction after release. $x_1 = 0 + 5t  0$ The cart’s initial horizontal velocity is 5 m/s, but it slows due to a constant acceleration opposite its initial direction of motion. $x_2 = 0 + 5t + \dfrac{1}{2}(1) t^2$ projectile motion on a moving cart Last edited by skeeter; March 18th, 2018 at 04:16 AM. 

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